Question: Simplify the following expression: $y = \dfrac{3x^2- 1x- 24}{x - 3}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(3)}{(-24)} &=& -72 \\ {a} + {b} &=& &=& {-1} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-72$ and add them together. Remember, since $-72$ is negative, one of the factors must be negative. The factors that add up to ${-1}$ will be your ${a}$ and ${b}$ When ${a}$ is ${8}$ and ${b}$ is ${-9}$ $ \begin{eqnarray} {ab} &=& ({8})({-9}) &=& -72 \\ {a} + {b} &=& {8} + {-9} &=& -1 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({3}x^2 +{8}x) + ({-9}x {-24}) $ Factor out the common factors: $ x(3x + 8) - 3(3x + 8)$ Now factor out $(3x + 8)$ $ (3x + 8)(x - 3)$ The original expression can therefore be written: $ \dfrac{(3x + 8)(x - 3)}{x - 3}$ We are dividing by $x - 3$ , so $x - 3 \neq 0$ Therefore, $x \neq 3$ This leaves us with $3x + 8; x \neq 3$.